Are the coefficients of $n^{th}$ order monic polynomial with distinct roots in $\mathbb C$ dense in $\mathbb R^n$

Question

Suppose for a $n^{th}$ order monic polynomial $p(t) = t^n + \alpha_{n-1}t^{n-1} + \dots + \alpha_0$, we identify the coefficients in $\mathbb R^n$, i.e., $\alpha = (\alpha_{n-1} ,\dots, \alpha_0) \in \mathbb R^n$. Let $$\mathcal F = \{(b_{n-1}, \dots, b_0): p(t)=t^{n} + b_{n-1} t^{n-1} + \dots + b_0 \text{ has $n$ distinct roots in $\mathbb C$}\}.$$ I am wondering whether $\mathcal F$ is dense in $\mathbb R^n$.

Answer 1

Yes.

Whether or not a polynomial has multiple roots can be read off its discriminant, which is a polynomial in the coefficients. The non-roots of any non-zero polynomial are dense.

A direct approach for $n>1$ (the case $n=1$ being clear) is to note that the multiple roots $f$ re the roots of $\gcd(f,f')$ and that polynomials differing by a non-zero constant have no roots in common. Hence already among all polynomials $f(x)+c$ with the same derivative $f'$, there can be only finitely many with $\gcd(f(x)+c,f'(x))\ne1$.

Answer 2

Yes, since asserting that $p(x)$ has $n$ distinct roots is equivalent to asserting that $\gcd\bigl(p(x),p'(x)\bigr)\neq1$ and this condition can be expressed by the fact that the coefficients of $p(x)$ are the non-zeros of a certain non-null polynomial.